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4

Following √Člie Cartan, you want to think of your differential system $dR - R\omega = 0$ on $M\times SO(n)$. This $\mathfrak{so}(n)$-valued $1$-form is integrable, as you said, because of vanishing curvature. The integral manifolds of this differential system will locally be the graphs of functions $R\colon U\to SO(n)$.


3

$$J=\int_{\mathbb{R}^n}(x_1^2+\ldots+x_n^2) e^{-\|x\|^2/4}\,d\mu = n\int_{\mathbb{R}^n}x_1^2 e^{-x_1^2/4}e^{-(x_2^2+\ldots+x_n^2)/4}\,d\mu $$ by symmetry. Since $\int_{-\infty}^{+\infty}x^2 e^{-x^2/4}\,dx = 4\sqrt{\pi}$ and $\int_{-\infty}^{+\infty}e^{-x^2/4}\,dx = 2\sqrt{\pi}$, by Fubini's theorem it follows that: $$ (4\pi)^{-n/2}\int_{\mathbb{R}^n}\|x\|^2 ...


2

(I would have just commented, but I don't yet have enough reputation) The solution is $u = xy$. Uniqueness tells us that this is the only solution. The "quickest" way (if you have been doing this for a little bit) is by inspection, which is a fancy name for an educated guess. You get better at doing this with simple PDE's as you work more with pdes. ...


2

You have $Lu = u_x + u u_y$, hence: $$L(a u + b v ) = a u_x + b v_x + (a u + b v) \, (au_y + bv_y)$$ Can you take it from here?


2

$$T=\partial/\partial x + I \partial/\partial y$$ where I is the identity operator. Applying it to a sum: $$T(u+v)=u_x+v_x+u u_y+uv_y+vu_y+vv_y=T(x)+T(y)+uv_y+vu_y;$$ as you can see you get two terms too many!


2

GENERAL DEVELOPMENT: First we establish the equivalence of the two forms for the Navier-Stokes Equations given in the OP. To do this, we use straightforward product rule differentiation to show that $$\begin{align} \frac{\partial \rho \vec v}{\partial t}=\frac{\partial \rho }{\partial t}\vec v+\rho \frac{\partial \vec v }{\partial t} \tag 1 \end{align}$$ ...


2

They are both correct. Remember you also have the divergence free condition: $$ \frac{\partial v}{\partial z} = 0 .$$ Note, this makes the 1D incompressible NS equation quite boring.


2

Both are differential equations (equations that involve derivatives). ODEs involve derivatives in only one variable, whereas PDEs involve derivatives in multiple variables. Therefore all ODEs can be viewed as PDEs. PDEs are generally more difficult to understand the solutions to than ODEs. Basically every big theorem about ODEs does not apply to PDEs. It's ...


2

Well, given a linear ODE, the set of solutions form a vector space with finite dimension. However, a linear PDE (like the heat equations) has a set of solution that form a vector space with infinitely many dimensions. To see that, one may consider the ODE $$ y'=-ay(t), $$ with solution, $$ y(t)=e^{-at}y_0, $$ (so, the vector space is one dimensional) ...


2

The proof looks correct to me, well done! A couple of nitpicks: in the definition of $V \subset \subset U$ you forgot to include that $\overline{V}$ needs to be compact; the very last $\delta$ should be $\delta^p$. The only thing that I would like to add is an explicit definition for $V_k$: $$V_k := \Big\{x \in U : \text{dist}(x, \partial U) > ...


2

You can use linearity like this: If $Lu_1 = g \tag{1}$ and $Lu_2 = g, \tag{2}$ then $Lu_1 - Lu_2$ $= g - g = 0; \tag{3}$ but, by linearity, $Lu_1 - Lu_2$ $= L(u_1 - u_2), \tag{4}$ so $L(u_1 - u_2) = 0, \tag{5}$ as per request.


2

Hints: Write the partial derivatives as operators, as follows: $$ u_{xx} = \frac{\partial^2}{\partial x^2} u $$ $$ u_{xt} = \frac{\partial^2}{\partial x\,\partial t} u $$ $$ u_{tt} = \frac{\partial^2}{\partial t^2} u $$ so the PDE looks like this: $$ \left(\,3\frac{\partial^2}{\partial x^2} - 10\frac{\partial^2}{\partial x\,\partial t} - ...


2

HINT: Note that we have $$\frac{\partial g}{\partial \nu}=\frac{2}{\epsilon}$$ on $\Gamma_{\epsilon}$. Therefore, $$\lim_{\epsilon\to 0}\int_{\Gamma_{\epsilon}}\left(g\frac{\partial\phi}{\partial\nu}-\phi\frac{\partial g}{\partial\nu}\right)\,ds=-4\pi\phi(0)$$


2

Check this link to see how to deal with differentiating an integral over an evolving domain: https://en.wikipedia.org/wiki/Time_evolution_of_integrals. We use this to see that $$\partial_t\left(\int_0^tf(x+(s-t)b,s)\,ds\right)=\int_0^t\frac{\partial f(x+(s-t)b,s)}{\partial t}\,ds+f(x,t),$$ now notice that $$\int_0^t\frac{\partial f(x+(s-t)b,s)}{\partial ...


1

Try $\frac{u(x+he_i)-u(x)}{h}=\int_{\mathbb{R}^n}\Phi(y)\{\frac{f(x+he_i-y)-f(x-y)}{h}\}dy$. Then $\frac{f(x+he_i-y)-f(x-y)}{h}\rightarrow\frac{\partial f}{\partial x_i}(x-y)$ uniformly(!) in $h$ and of course $u\in C^1$ implies $u\in C^0$. Similarly You can show $u\in C^2$.


1

If $R$ is large enough $\varphi$ and $\nabla \varphi$ are equal to $0$ on $\Gamma_2$, and so is the RHS in the last formula.


1

The first equation looks like the Eikonal equation in $\Bbb R^{1+1}$, it is solved by $u=\sqrt{x-t}$, you would then plug this into the wave equation to determine $f$. We see that $u_x=x/u$, $u_t=-t/u$, $u_{xx}=(u-x^2/u)/u^2$ and $u_{tt}=-(u+t^2/u)/u^2$ thus$$u_{xx}-u_{tt}=(u-x^2/u)/u^2+(u+t^2/u)/u^2$$ $$=2u/u^2+(t^2-x^2)/u^3$$ ...


1

You can think of a linear functional on a vector space as giving you a coordinate in that space. A continuous linear functional is a coordinate that varies continuously as the vector varies continuously; one is not so much interested in the coordinates that are disconnected from the topology of the space. Representation of a linear functional gives the ...


1

I think the problem is more a problem of basic proof-strategy/proof-writing than really a problem about linearity or differential equations. Therefore, I will try to show how mechanical this kind of proof can be. An interesting lecture on that topic is this blogpost by the Field medalist Tim Gowers. The difference between two solutions of an ...


1

Taking the first order partial derivatives gives $$z_x=2xf'(x^2+y^2)+1\\ z_y=2yf'(x^2+y^2)+1 $$ can you relate these two expressions somehow?


1

You are pretty much there. Since $f$ is continuous on $[0,M]$, it follows that $\lim_{\varepsilon\to0^+}\int_\varepsilon^Mf(x)\, dx=\int_0^Mf(x)\, dx$. EDIT: $f$ here is defined as you originally did before you edited the question: $$f(x):=\begin{cases} \frac{\varphi(x)-\varphi(0)}x&\text{if }x>0,\\ \varphi'(0)&\text{if }x=0. \end{cases}$$


1

Note that $\int_{\mathbb{R}} |\nabla u(x)|^2 dx=-\int_{\mathbb{R}}u(x)\Delta u(x)dx=0$. So $\nabla u(x)=0$ and $u$ is identically a constant.


1

Perhaps it would be wiser to use the following expansion $$E(f+h)=E(f)+dE(h)+o(h),$$ from what you have calculated, we see that $$\int \nabla f\cdot\nabla h+\frac{1}{2}\nabla h\cdot\nabla h=E(f+h)-E(f)=dE(h)+o(h).$$ Since the the last term on the LHS is $o(h)$, we obtain $dE(h)=\int\nabla f\cdot\nabla h$. Note that this is the Frechet derivative at the ...


1

I think it would be best to send the authors an email and ask them, since they are the only ones who knows by certain. Nevertheless, let me make a guess: I think that $$ u\wedge M=\min(u,M) $$ and that $$ u\vee M=\max (u,M). $$ I'm not certain of what they want to do, but I guess that they are cutting the sequence $\phi_n$ off at the values $\pm M$ (in ...


1

You are correct. A measure has no reason to preserve this property. As a matter of fact, the only measures that absolutely continuous wrt Lebesgue measure that preserve this property for all $f,g \in C^1_c(\Bbb{R})$ are scaled Lebesgue measures. Let $g\neq 0$. If $(gh)'=g'h$, then $gh'+g'h=g'h$, or $ gh'=0$, implying $h'=0$ and $h=c$. Or if the question ...


1

Let $u=v^2$ , Then $u_x=2vv_x$ $u_y=2vv_y$ $\therefore2vv_x2vv_y=v^2$ with $v(x,0)=0$ $4v^2v_xv_y=v^2$ with $v(x,0)=0$ $v_y=\dfrac{1}{4v_x}$ with $v(x,0)=0$ $v_{xy}=-\dfrac{v_{xx}}{4v_x^2}$ with $v(x,0)=0$ Let $w=v_x$ , Then $w_y=-\dfrac{w_x}{4w^2}$ with $w(x,0)=0$ $\dfrac{w_x}{4w^2}+w_y=0$ with $w(x,0)=0$ Follow the method in ...


1

Linear PDEs may not have solutions. Hans Lewy constructed such an example about sixty years ago, and it probably surprised just about everyone in the field at the time. ODEs are much nicer in that regard. http://www.jstor.org/stable/1970121?&seq=1#page_scan_tab_contents


1

The methodology in the posted question was correct and gives a way forward. Its implementation had some mistakes which we resolve here. We begin with $$\frac{\partial^2 f(x,y)}{\partial x\partial y}=e^{x+2y} \tag 1$$ and integrate $(1)$ with respect to $x$ to obtain $$\frac{\partial f(x,y)}{\partial y}=e^{x+2y}+C_1(y) \tag 2$$ where $C_1(y)$ is an ...



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